Solve the system of equations
$6 (a x + b y) = 3 a + 2 b$
$6 (b x - a y) = 3 b - 2 a$

\frac{1}{3}, \frac{1}{2}

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\frac{1}{6}, \frac{2}{3}

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\frac{- 1}{3}, \frac{1}{6}

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\frac{1}{2}, \frac{1}{3}

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Solution

The correct option is C $\frac{1}{2}, \frac{1}{3}$
Rearranging the equations
$6 a x + 6 b y = 3 a + 2 b$
$6 b x - 6 a y = 3 b - 2 a$
comparing the coefficeint (sonsotants) on LHS
and RHS of the equation $(1)$ say it is very clear that option D can only fit

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Similar questions

6 (a x + b y) = 3 a + 2 b, 6 (b x - a y) = 3 b - 2 a .

6 (a x + b y) = 3 a + 2 b

6 (b x - a y) = 3 a - 2 b

6 (a x + b y) = 3 a + 2 b 6 (b x - a y) = 3 b - 2 a

Solve for x and y:

$6 (a x + b y) = 3 a + 2 b, 6 (b x - a y) = 3 b - 2 a .$

2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5, (x \neq - 3, \frac{1}{2})

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